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2.643 grams of potassium butanoate (KCH3(CH2)2CO2 ) is fully dissolved in 50.00 mL of water, which is carefully transferred to a conical flask. Then 100.00 mL of 0.120 M HCℓ is added dropwise to this solution from a burette. Given: Ka(butanoic acid) = 1.5 × 1O−5 . 2.1 Showing all your calculations and reasoning, determine the pH of the solution that results after the addition of all the acid mentioned above.

Respuesta :

Answer:

The pH of the solution is 4.69

Explanation:

Given that,

Mass of potassium = 2.643 grams

Weight of water = 50.00 mL

Weight of HCl=100.00 ml

Mole = 0.120 M

We know that,

[tex]KCH_{3}(CH_{2})_{2}CO_{2}[/tex] is a basic salt.

Let's write it as KY.

The acid [tex]HCH_{3}(CH_{2}CO_{2})[/tex] would become HY.

We need to calculate the moles of KY

Using formula of moles

[tex]moles\ of\ KY=\dfrac{m}{M}\times1000[/tex]

[tex]moles\ of\ KY=\dfrac{2.643}{126}\times1000[/tex]

[tex]moles\ of\ KY=20.97\ m\ mole[/tex]

The reaction is

[tex]KY+HCl\Rightarrow HY+ KCl[/tex]

The number of moles of KY is 20.98 m

initial moles = 20.98

Final moles [tex]m=20.98-0.120\times100= 8.98[/tex]

We need to calculate the value of pKa(HY)

Using formula for pKa(HY)

[tex]pKa_{HY}=-log Ka[/tex]

[tex]pKa_{HY}=-log(1.5\times10^{-5})[/tex]

[tex]pKa_{HY}=4.82[/tex]

We need to calculate the pH of the solution

Using formula of pH

[tex]pH=pKa+log(\dfrac{[KY]}{[KH]})[/tex]

Put the value into the formula

[tex]pH=4.82+log(\dfrac{8.98}{12})[/tex]

[tex]pH=4,69[/tex]

Hence, The pH of the solution is 4.69

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